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Question

Evaluate x2+1x4x2+1dx

A
tan1(x2+1x)+c
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B
cot1(x2+1x)+c
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C
cot1(x21x)+c
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D
tan1(x21x)+c
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Solution

The correct option is D tan1(x21x)+c
Let I=x2+1x4x2+1dx=1+1x2x21+1x2dx=1+1x2(x1x)2+1dx
Put x1x=t

(1+1x2)dx=dt
I=dtt2+1

=tan1t+c ..................... 1a2+x2dx=1atan1(xa)+c

=tan1(x1x)+c

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